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3 years ago

The equation of a circle is x2+4x+y2−10y+13=0 .

Mathematics

2 answers:

kiruha [24]3 years ago

8 0

X^2+4x+4+y^2-10y+25=-13+25+4

(x+2)^2 + (y-5)^2= 16
Third option

Afina-wow [57]3 years ago

6 0

Answer:

$(x+2)^2+(y-5)^2=16$

Step-by-step explanation:

Given : $x^2+4x+y^2-10y+13=0$

To Find: What is the equation of the circle in standard form?

Solution:

$x^2+4x+y^2-10y+13=0$

$x^2+4x+2^2-2^2+y^2-10y+5^2-5^2+13=0$

$(x+2)^2-4+(y-5)^2-25+13=0$

$(x+2)^2+(y-5)^2-29+13=0$

$(x+2)^2+(y-5)^2-16=0$

$(x+2)^2+(y-5)^2=16$

Hence the equation of the circle in standard form is $(x+2)^2+(y-5)^2=16$

So, Option C is correct.

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2 years ago

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Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$